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Question: Prove $\epsilon - \delta$ style that $\lim\limits_{x \rightarrow 2}x^2 \neq 6$ via contradiction

So my initial idea is to assume $\lim\limits_{x \rightarrow 2} x^2 = 6$. Then for all $\epsilon > 0$ $\exists$ $\delta > 0$ such that $|x^2-6| < \epsilon \rightarrow0 < |x-2| < \delta$

However, I am not sure how to show a contradictio without "plugging it in".... could someone show me?

RobPratt
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A. Radek Martinez
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    The structure of the proof is that you would, beforehand, find a $\delta, \epsilon$ pair that fail. Then your proof would say "assume this is the limit. Then we have this inequality. However, this pair of $\delta, \epsilon$ does not satisfy this inequality. Therefore our assumption was false". It's basically the classic way of making a direct proof harder by trying to prove it by using contradiction. – Osama Ghani Aug 19 '20 at 00:53
  • Can you invoke uniqueness of limits? If so, you could show the limit is $4$ and use $4\ne 6$. – Integrand Aug 19 '20 at 02:07
  • https://math.stackexchange.com/questions/95026/prove-that-the-limit-does-not-go-to-6 – C Squared Aug 19 '20 at 02:30

1 Answers1

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Let $\varepsilon = 0.25 > 0 $

We have that for all $\delta > 0$, if we take $\alpha = \text{min}\{0.1,\frac{\delta}{2} \}$, then we have that $2\alpha + \alpha^{2} \leq 0.2 + 0.01 = 0.21$

If we take $x = 2 + \alpha$, we have that $|2+ \alpha - 2| = \alpha < \delta$, but $|(2+ \alpha)^{2} - 6| = 6 - 4 - 2\alpha - \alpha^{2} \geq 2 - 0.21 > 1 > \varepsilon$. Then is a contradiction of the limit definition

ZAF
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